As air pressure decreases, what happens

What is the average speed of a car that travelled 400 miles in 6 hours?

Reil [10]

Answer:

About 66 miles per hour

Explanation:

Based on the information given we can assume the car traveled the same number of miles every hour meaning all we need to do is divide.

400/6 ≈ 66 miles per hour

8 0

3 years ago

Read 2 more answers

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

PHYSICS, HELP PLZ??!! 100PTS

romanna [79]

Hi there!

We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:

$t_{max} = \frac{vsin\theta}{g}$

Where:

tmax = (? sec)

vsinθ = vertical comp. of velocity = 10sin(48) = 7.43 m/s)

g = acceleration due to gravity (9.8 m/s²)

We can solve for this time:

$t_{max} = \frac{7.43}{9.8} = 0.758 s$

When the ball is at the TOP of its trajectory, its VERTICAL velocity is equivalent to 0 m/s. Thus, we can consider this a free-fall situation.

We must begin by solving for the maximum height reached by the ball using the equation:

$d = y_0 + v_{0y}t + \frac{1}{2}at^2$

d = displacement (m)

vi = initial velocity (7.43 m/s)

a = acceleration due to gravity

d = displacement (m)

y0 = initial VERTICAL displacement (28m)

Plug in the values:

$d = 28 + 7.43(0.758) + \frac{1}{2}(-9.8)(0.758^2) = 30.817 m$

Now, we can use the rearranged kinematic equation:

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(30.817)}{9.8}} = 2.51 s$

Add the two times together:

$0.758 + 2.51 = \boxed{3.266 s}$

7 0

2 years ago

What types of crust are colliding between the South American Plate and the Nazca Plate?

Neko [114]

Answer:

Oceanic Crust/Continental Crust

Explanation:

It is a destructive plate boundary between the oceanic crust of the Nazca plate and the continental crust of the South American plate.

3 0

3 years ago

What is the period and group of gold

algol13

Answer:

Chemical elements, a dense lustrous yellow precious metal of group.

4 0

3 years ago